Question: $f(x) = 7x^{3}+4x^{2}-7x-1-2(g(x))$ $h(n) = -6n^{3}+5n^{2}-3n+3+f(n)$ $g(t) = -6t$ $ g(f(1)) = {?} $
First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = 7(1^{3})+4(1^{2})+(-7)(1)-1-2(g(1))$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = (-6)(1)$ $g(1) = -6$ That means $f(1) = 7(1^{3})+4(1^{2})+(-7)(1)-1+(-2)(-6)$ $f(1) = 15$ Now we know that $f(1) = 15$ . Let's solve for $g(f(1))$ , which is $g(15)$ $g(15) = (-6)(15)$ $g(15) = -90$